Question

What is the formula of $x^3+y^3+z^3-3xyz$?

Answer 1

Write the formula for$x^3+y^3+z^3-3xyz$.

Add and subtract expressions $3x^{2}y\mathrm{and}3x{y}^2$ in $x^3+y^3+z^3-3xyz$.

$$\begin{gathered} x^3+y^3+z^3-3xyz=x^3+y^3+3x^{2}y+3x{y}^2+z^3-3xyz-3x^{2}y-3x{y}^2 \\ {x}^3+y^3+z^3-3xyz={\left(x+y\right)}^3+z^3-3xy\left(x+y+z\right)\left[\mathbf{\because }{\left(x+y\right)}^{\mathbf{3}}\mathbf{=}{\mathit{x}}^{\mathbf{3}}\mathbf{+}{\mathit{y}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{+}\mathbf{3}\mathit{x}{\mathit{y}}^{\mathbf{2}}\right] \end{gathered}$$

Use formula${\mathit{u}}^{\mathbf{3}}\mathbf{+}{\mathit{v}}^{\mathbf{3}}\mathbf{=}\left(u+v\right)\left(u^2+v^2-uv\right)$.

Put $u=x+y,v=z$in the formula:

$$\begin{gathered} x^3+y^3+z^3-3xyz=\left[\left(x+y\right)+z\right]\left[{\left(x+y\right)}^2+z^2-\left(x+y\right)z\right]-3xy\left(x+y+z\right) \\ \Rightarrow x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left[{\left(x+y\right)}^2+z^2-xz-yz\right]-3xy\left(x+y+z\right)\left[\mathbf{\because }{\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{)}}^{\mathbf{2}}\mathbf{=}\left(x^2+y^2+2xy\right)\right] \\ \Rightarrow x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+2xy+z^2-xz-yz\right)-3xy\left(x+y+z\right) \\ \Rightarrow x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+2xy+z^2-xz-yz-3xy\right) \\ \Rightarrow x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xz-yz-xy\right) \end{gathered}$$

Hence, $\begin{array}{rcl}{x}^3+y^3+z^3-3xyz& =& \left(x+y+z\right)\left(x^2+y^2+z^2-xz-yz-xy\right)\end{array}$